On the Diophantine equation k

On the Diophantine equation k j=1 jfp j = Fq n arxiv:1705.06066v1 [math.nt] 17 May 017 Gökhan Soydan 1, László Németh, László Szalay 3 Abstract Let F n denote the n th term of the Fibonacci sequence. Inthis paper, we investigate the Diophantine equation F p 1 +Fp + +kfp k = Fq n in the positive integers k and n, where p and q are given positive integers. A complete solution is given if the exponents are included in the set {1,}. Based on the specific cases we could solve, and a computer search with p,q,k 100 we conjecture that beside the trivial solutions only F 8 = F 1 +F +3F 3 +4F 4, F4 = F 1 +F +3F 3, and F4 3 = F3 1 +F3 +3F3 3 satisfy the title equation. Key Words: Fibonacci sequence, Diophantine equation. AMS Classification: 11B39, 11D45. 1 Introduction As usual, let (F n ) n 0 and (L n ) n 0 denote the sequences of Fibonacci and Lucas numbers, respectively, given by the initial values F 0 = 0, F 1 = 1, L 0 =, L 1 = 1, and by the recurrence relations F n+ = F n+1 +F n and L n+ = L n+1 +L n for all n 0, (1) respectively. Putting α = (1 + 5)/ and β = (1 5)/ = 1/α for the two roots of the common characteristic equation x x 1 = 0 of the two sequences, the formulae F n = αn β n α β and L n = α n +β n hold for all n 0. These numbers are well-known for possessing amazing and wonderful properties (consult, for instance, [13] and [5] together with their very rich annotated bibliography 1 Department of Mathematics, Uludağ University, Görükle Kampüs, 16059 Bursa, Turkey. gsoydan@uludag.edu.tr University of Sopron, Institute of Mathematics, H-9400, Sopron, Bajcsy-Zs. utca 4., Hungary. nemeth.laszlo@uni-sopron.hu 3 University of Sopron, Institute of Mathematics, H-9400, Sopron, Bajcsy-Zs. utca 4., Hungary; J. Selye University, Institute of Mathematics and Informatics, 94501, Komárno, Bratislavska cesta 33, Slovakia. szalay.laszlo@uni-sopron.hu 1

G. Soydan, L. Németh, L. Szalay for history and additional references). Observing F 1 = F, F 1 +F = F 4, F 1 +F +3F 3 = F 4, F 1 +F +3F 3 +4F 4 = F 8, the question arises naturally: is there any rule for F 1 +F +3F 3 + +kf k? We study this question more generally, according to the title equation. Diophantine equations among the terms of Fibonacci numbers have a very extensive literature. Here we quote a few results that partially motivated us. By the defining equality (1) of the Fibonacci numbers and the identity Fn +F n+1 = F n+1 (Lemma 1.8), we see that Fn s + Fn+1 s (n 0) is a Fibonacci number for s {1,}. For larger s Marques and Togbé [8] proved in 010 that if Fn s + Fs n+1 is a Fibonacci number for all sufficiently large n then s = 1 or. Next year Luca and Oyono [7] completed the solution of the question by showing that apart from F1 s + F s = F 3 there is no solution s 3 to the equation Fn s +Fs n+1 = F m. Letl,s 1,...,s l,a 1,...,a l beintegerswithl 1ands j 1. Supposethatthereexists 1 t l such that a t 0 and s t > s j, for all j t. Chaves, Marques and Togbé [4], showed that if either s t is even or a t is not a positive power of 5, then the sum a 1 F s 1 n+1 +a F s n+ +...+a l F s l n+l does not belong to the Fibonacci sequence for all sufficiently large n. A balancing problem having similar flavor has been considered by Behera et al. [3]. They studied the equation F p 1 +F p + +F p k 1 = Fq k+1 + +Fq k+r, () and solved it for the cases (p,q) = (,1),(3,1),(3,), and for p q by showing the nonexistence of any solution. Further the authors conjectured that only the quadruple (k, r, p, q) = (4, 3, 8, ) of positive integers satisfies (). The conjecture was completely justified by Alvarado et al. [1]. Note that if (p,q) = (1,1) we obtain the problem of sequence balancing numbers handled by Panda [9]. Recalling the formulae F 1 +F + +F k = F k+ 1 and F 1 +F + +F k = F kf k+1, it is obvious that the problems F 1 +F + +F k = Fn q, and F 1 +F + +F k = Fq n are rather simple. Indeed, the equations above lead to the lightsome ones F k+ 1 = Fn, q F k F k+1 = Fn. q However the equation F p 1 +F p + +F p k = Fq n might be taken an interest if p 3. The last motivation of our examination was the Diophantine equation x +(x+1) + +n(x+n 1) = y (3)

On the Diophantine equation k j=1 jfp j = Fq n 3 to determine the values of n for which it has finitely or infinitely many positive integer solutions (x,y) (see Wulczyn [14], and for details, see also []). For variations of the equation (3), we refer the reader to [1]. In this paper, we investigate the Diophantine equation F p 1 +F p + +kf p k = Fq n (4) in the positive integers k and n, where p and q are fixed positive integers. We consider F p 1 = 1 = Fq 1 = Fq, and Fp 1 +Fp = 3 = F 4 as trivial solutions to (4). We have the following conjecture based upon the specific cases we could solve, and a computer search with p,q,k 100. Conjecture 1. The non-trivial solutions to (4) are only F 4 = 9 = F 1 +F +3F 3, F 8 = 1 = F 1 +F +3F 3 +4F 4, F 3 4 = 7 = F 3 1 +F3 +3F3 3. This work handles the particular cases p,q {1,} (hence the first two solutions above will be obtained), the precise results proved are described as follow. Theorem 1. If F 1 +F + +kf k = F n, (5) then (k,n) = (1,1), (1,), (,4), (4,8), among them only the last one is non-trivial solution. Theorem. The Diophantine equation F1 +F + +kf k = F n (6) possesses only the trivial solutions (k,n) = (1,1), (1,). Theorem 3. If F 1 +F + +kf k = F n, (7) then (k,n) = (1,1), (1,), (3,4), among them only the last one is non-trivial solution. Theorem 4. The Diophantine equation F1 +F + +kfk = F n (8) possesses only the trivial solutions (k, n) = (1, 1), (1, ), (, 4).

4 G. Soydan, L. Németh, L. Szalay Lemmata In this section, we present the lemmata that are needed in the proofs of the theorems. The first lemma is a collection of a few well-known results, we state them without proof, and in the proof of the theorems sometimes we do not refer to them. Lemma 1. Let k and n be arbitrary integers. 1. k j=1 jf j = kf k+ F k+3 +.. k j=1 jf j = F k (kf k+1 F k )+τ, where τ = 0 if k is even, and τ = 1 otherwise. 3. For k 0 we have F k = ( 1) k+1 F k, further L k = ( 1) k L k (extension of the sequences for negative subscripts). 4. gcd(f k,f n ) = F gcd(k,n). 5. gcd(f k,l n ) = 1 or or L gcd(k,n). 6. F k F n if and only if k n. 7. F k+1 F n F k F n+1 = ( 1) n+1 F k n (d Ocagne s identity). 8. F k+n = F k F n+1 +F k 1 F n. 9. F k = F k L k. 10. F k+n F k n = F kl n. Lemma. F (k+)/ L (k )/, if k 0 (mod 4) F (k 1)/ L (k+1)/, if k 1 (mod 4) F k 1 = ; F (k )/ L (k+)/, if k (mod 4) F (k+1)/ L (k 1)/, if k 3 (mod 4) F (k )/ L (k+)/, if k 0 (mod 4) F (k+1)/ L (k 1)/, if k 1 (mod 4) F k +1 =. F (k+)/ L (k )/, if k (mod 4) F (k 1)/ L (k+1)/, if k 3 (mod 4) Proof. See, for instance, [6] and [10]. Lemma 3. F k 1 = { F k 1 F k+1, if k 1 (mod ) F k F k+, if k 0 (mod ) ; F k +1 = { F k 1 F k+1, if k 0 (mod ) F k F k+, if k 1 (mod ).

On the Diophantine equation k j=1 jfp j = Fq n 5 Proof. See Lemma 3 in [11]. Lemma 4. If j 4 is even, then Proof. Since gcd(f j 1,F j ) = 1, and F ϕ(f j) j 1 But F ϕ(f j) 1 j 1 F j 3 (mod F j ). 1 (mod F j ), it is sufficient to show that F j 3 F j 1 (mod F j ). F j 1 F j 3 = (F j+1 F j )F j 3 F j+1 F j 3 = F j F j +( 1) j F 3 (mod F j ) follows from the definition of the Fibonacci numbers, d Ocagne s identity (Lemma 1.7), and the parity of j. Lemma 5. If j 3 is odd, then F ϕ(f j) 1 j 1 F j (mod F j ). Proof. Similarly to the proof of the previous lemma, the statement is equivalent to And it is easy to see that 1 F j F j 1 (mod F j ). F j F j 1 = (F j F j 1 )F j 1 F j 1 = (F j F j +( 1) j F 1 ) 1 (mod F j ). Lemma 6. Let k 0 be a positive integer, and for i {0,1} put ( ) 1+( 1) i 1 ( β /α) k 0 δ i = log α, 5 where log α is the logarithm in base α = (1 + 5)/. Then for all integers k k 0, the two inequalities α k+δ 0 F k α k+δ 1 hold. Proof. This is a part of Lemma 5 in [6]. In order to make the application of Lemma 6 more convenient, we shall suppose that k 0 1. Then we have Corollary 5. If k 1, then α k F k α k 1, and equality holds if and only if k =, and k = 1, respectively. Now, we are ready to justify the theorems.

6 G. Soydan, L. Németh, L. Szalay 3 Proofs Proof of Theorem 1. Verifying the cases k = 1,...,5 by hand we found the solutions listed in Theorem 1. Put = k+, and suppose that 8. Consequently, F 3 5 and F 1. If equation (5) holds, then n >, and then by Lemma 1.1 we conclude k = F n +F +1 F = F n +F 1 F +1 N. (9) In the sequel, we study the sequence (F u ) u=0 modulo F if is fixed. Note that we indicate a suitable value congruent to F u modulo F, not always the smallest non-negative remainders. The period can be deduced from the range {}}{ 0,1,1,,...,F,F 1, {}}{ 0,F 1,F 1,F 1,...,F F 1,F 1 F 1, of length if is even, since then, by Lemma 1.7 we have F 1 1 (mod F ) and then F F 1 = (F F 1 )F 1 1 (mod F ). In case of odd we have F 1 1 (mod F ), therefore the length of the period is 4 coming from {}}{{}}{ 0,1,1,,...,F,F 1, 0,F 1,F 1,F 1,...,F F 1,F 1 F 1, {}}{{}}{ 0, 1, 1,,..., F, F 1, 0, F 1, F 1, F 1,..., F F 1, F 1 F 1. Based on the length of the period we distinguish two cases. Case I: is even. Either F n F j or F n F j F 1 modulo holds for some j = 0,1,..., 1. Hence { F j +F 1, or F n +F 1 (mod F ). (10) F j F 1 +F 1 We will show that none of them is congruent to 0 modulo F. In the first branch further if j 1, then F j +F 1 F 1 11, F j +F 1 F +F 1 F. Thus F j + F 1 0 (mod F ), hence (9) does not hold. Assume now, that j = 1. Then, together with the definition of the Fibonacci sequence we have F j +F 1 = F 1 +(F F ) F 3 (mod F ).

On the Diophantine equation k j=1 jfp j = Fq n 7 But 3 F 3 < F contradicts to (9). Choosing the second branch of (10), suppose that F 1 (F j +1) is congruent to 0 modulo F. Then F ϕ(f) 1 (F j +1) F ϕ(f) 1 1 (mod F ). Subsequently, by Lemma 4, it leads to F j +1 F 3 (mod F ). Since j = 0,1,..., 1, ( 8) it follows that F j = F 3 1, a contradiction. Case II: is odd. Now 9, and either F n ±F j (mod F ) holds for some j = 0,1,..., 1. Hence { ±F j +F 1 F n +F 1 ±F j F 1 +F 1 (mod F ) or F n ±F j F 1 (mod F ). First, obviously, if j 1, then 6 F 3 ±F j +F 1 F, so dividing ±F j +F 1 by F, the result is not an integer. If j = 1, then the treatment of the + casecoincidesthetreatmentwhenwaseven. The caseleadstof n +F 1 (mod F ), a contradiction. Assume now that F n +F 1 ±F j F 1 +F 1 (mod F ). Thus F 1 (1±F j ) (mod F ). Multiplying both sides by F ϕ(f) 1 1, by Lemma 5 it gives 1±F j F (mod F ). First let F j = F 1, which leads immediately a contradiction via 0 < F 1 = F 1 + F 4 1 < F. If F F j + 1 = F, then F j = F 3 + 1 follows, a contradiction again. The proof of Theorem 1 is complete. Proof of Theorem. For the range k = 1,,...,0 we checked (6) by hand. From now we assume k 1. Based on Lemma 1., we must distinguish two cases. Case I: k is even. Consider the equation F k (kf k+1 F k ) = F n. Trivially, n > k. Put ν = gcd(k,n). If ν = k, then F k F n by Lemma 1.6. Consequently, ( Fn F k ) = kf k+1 F k = kf k+1 1 F k F k is integer. But F k and F k+1 are coprime, hence F k k, and it results k 5, a contradiction.

8 G. Soydan, L. Németh, L. Szalay Examine the possibility ν = k/. Put = k/. Now F L (kf k+1 F L ) = Fn leads to ( ) L (kf k+1 F L ) Fn =. F This is an equality of integers, which together with gcd(f,f k+1 ) and gcd(f,l ) = 1, (see Lemma 1.5) shows that k/f is integer. Thus k 14, a contradiction. Finally, we have 3 ν k/3. Since gcd(f k /F ν,f n /F ν ) = 1, then from the equation F F k F ν (kf k+1 F k ) = F n F ν F n we conclude F k F ν F n and F n F ν kf k+1 F k. Subsequently, F k F ν F n and F n F ν (kf k+1 F k ). Thus F k F ν (kf k+1 F k ), and then F k kf ν holds since gcd(f k,f k+1 ) = 1. Applying Corollary 5, we obtain which implies k < 19. α k F k kf ν kα ν kα /3k, Case II: k is odd. In this part, we follow the idea of the previous case. Recall that k 1. Now F k (kf k+1 F k ) = F n ε F n+ε, where ε = 1 or depending on the parity of n (see Lemma 3). Clearly, gcd(n ε,n+ε) = or 4. Thus gcd(f n ε,f n+ε ) = 1 or 3, respectively. Put ν 1 = gcd(k,n ε) andν = gcd(k,n+ε). Obviously, gcd(ν 1,ν ) divides gcd(n ε,n+ε). Hence ν = gcd(ν 1,ν ) = 1 or or 4, and then F ν = gcd(f ν1,f ν ) = 1 or 3. Thus F ν1 F ν F ν F k. The terms of both the left and right sides of F k F ν1 (kf k+1 F k ) = F n ε F ν1 F n+ε and F k F ν (kf k+1 F k ) = F n ε F n+ε F ν are integers, and F n ε F ν1 kf k+1 F k, F k F ν F n ε. Combining them, F k F ν1 F ν (kf k+1 F k ) follows, and then F k kf ν1 F ν. The remaining part of the proof consists of three cases. Suppose first that ν 1 = k, i.e. F k F n ε. Observe, that n ε and n + ε are even, and k (n ε)/. Thus k +1 (n ε)/+1, which does not exceed (n+ε)/. Then kf k+1 > F n ε F ν1 F n+ε = F (n ε)/ F ν1 L (n ε)/ F n+ε L (n ε)/ L (n+ε)/ F (n+ε)/ L (n ε)/ L (n+ε)/ F k+1. Simplifying by F k+1 we conclude n ε k > L (n ε)/ L (n+ε)/,

On the Diophantine equation k j=1 jfp j = Fq n 9 and we arrived at a contradiction since 1 k < n. Note that the same machinery works when ν = k, i.e. F k F n+ε. If none of the two conditions above holds, we can assume ν 1 k/3 and ν k/3. Indeed, k is odd, so the largest non-trivial divisor of k is at most k/3. The application of Corollary 5 gives α k F k kf ν1 F ν kα ν 1 1 α ν 1 kα k/3, and then k < 19. The proof of the theorem is complete. Proof of Theorem 3. The proof partially follows the proof of Theorem 1. The small cases of (7) can be verified by hand. Suppose = k + 9. Similarly to (9), we have k = F n +F +1 F = F n +F 1 F +1 N. (11) Now we study the sequence (F u ) u=0 modulo F, and we again indicate the most suitable values by modulo F, not always the smallest non-negative remainders. Lemma 1.10, together with Lemma 1.3 implies F ±j F j (mod F ), where j = 0,1,..., 1. Hence the period having length can be given by {}}{{}}{ 0,1,1,,...,F,F 1, 0,F 1,F,...,,1,1. Let us distinguish two cases according to the parity of. Case I: is even. Put = l. Again by Lemma 1.10, together with i = l i = l + (l i) and i = l (l i) admits F i Fi (mod F ). It reduces the possibilities to j = 0,1,...,l. If j l 1 = ( )/, then 0 < F j +F 1 F ( )/ +F 1 F +F 1 < F hold since F ( )/ F ( )/L ( )/ = F. Suppose now that j = l = /. Repeating the previous idea we find F / + F 1 F +F 1 < F. Consequently, F / +F 1 = F mightbefulfilled. ThusF / = F. Recalling = l we equivalently obtain F l 1 = F l +1. Both sides have decomposition described in Lemma 3 and Lemma, respectively, providing F l 1 F l+1 = F l L l or F l F l+ = F l L l if l is odd or even, respectively. Firstly, F l F l 1 F l+1, and then F l F l+1 holds only for small l values. Secondly, F l F l F l+ contradicts to l 5. Case II: is odd. Now we have F i F i (mod F ). Indeed, Lemma 1.8 admits F i = (F F i+1 +F 1 F i ) F 1 F i (mod F ),

10 G. Soydan, L. Németh, L. Szalay and then Lemma 3 justifies the statement. It makes possible to split the proof into a few parts. If j = ( 1)/, then F( 1)/ +F 1 < F 1 < F. Thus F( 1)/ +F 1 = F is the only one chance to fulfill (11). Then apply Lemma 6 with k 0 = 4 ( 1)/ for F < F( 1)/ to reach a contradiction. If j ( 3)/, then 0 < F j +F 1 F ( 3)/ +F 1 F 3 +F 1 < F holds since F ( 3)/ F ( 3)/L ( 3)/ = F 3. Finally, if (+1)/ j 1, then F j +F 1 F j +F 1 (mod F ), where 1 J = j ( 1)/. Thus we need to check the equation FJ + = F 1, since FJ +F 1 < F 1 < F. When J ( 3)/ holds then FJ + < F 3+ < F 1. Lastly, J = ( 1)/ leads to F( 1)/ + = F 1, and then to = F J (L J F J ). Obviously, it gives J 3. Thus 7, a contradiction. Proof of Theorem 4. The statement for (8) is obviously true if k 1. So we may assume k 13. Let τ {0,1}. The formula of the summation, together with Corollary 5 implies α n F n = F k (kf k+1 F k )+τ < kf k F k+1 α log α k+k 1+(k+1) 1, and then n k < k +1+log α k < 3 k. Similarly, α n 1 F n = F k (kf k+1 F k )+τ > F k (kf k+1 F k+1 ) = (k 1)F k F k+1 > α log α (k 1)+k +(k+1), subsequently n k > k +log α (k 1) > k +3, that is n k k +4. Putting together the two estimates it gives k +4 n < 5 k. (1) Case I: k is even. Clearly, k+ < n k 1 3k/ holds. By Lemma 1.8, we conclude F n = F k+1 F n k +F k F n k 1 = kf k F k+1 F k, and equivalently F k (F k +F n k 1 ) = F k+1 (kf k F n k ). Thus gcd(f k,f k+1 ) = 1 admits F k+1 F k + F n k 1. The periodicity of (F u ) u=0 modulo F k+1 guarantees, together with the bounds on n k 1 that F k +F n k 1 F k +F j F k = F k (F j +1) (mod F k+1 )

On the Diophantine equation k j=1 jfp j = Fq n 11 holds for some j = 1,,...,k/ 3. Consequently, F k+1 F j +1, a contradiction. Case II: k is odd. Again k + < n k 1 3k/ holds. Lemma and Lemma 1.8 imply F k (kf k+1 F k ) = F (n ε)/ L (n+ε)/ = (F k+1 F (n ε)/ k +F k F (n ε)/ k 1 )L (n+ε)/, where ε {±1,±} according to the modular property of n. It leads to Thus F k F (n ε)/ k L (n+ε)/. By (1) it is obvious that F k+1 (kf k F (n ε)/ k L (n+ε)/ ) = F k (F k +F (n ε)/ k 1 L (n+ε)/ ). k < k +1 n n+ε n+ 5 k +1 < k, 4 which exclude gcd(k,(n+ε)/) = k. Thus gcd(k,(n+ε)/) k/3 since k is odd, subsequently gcd(f k,l (n+ε)/ ) L k/3. On the other hand n ε k n+ k 1 4 k +1, which implies gcd(f k,f (n ε)/ k ) F k/4+1. Thus, F k F (n ε)/ k L (n+ε)/, together with the previous arguments entails F k F k/4+1 L k/3, but it leads to a contradiction since L k/3 = F k/3 1 +F k/3+1 < F k/3+1, and the application of Corollary 5 on F k F k/4+1 F k/3+1 returns with k < 9. Acknowledgments. This paper was partially written when the third author visited the Department of Mathematics of Uludağ University in Bursa. He would like to thank the Turkish colleagues of the department for the kind hospitality. The first author was supported by the Research Fund of Uludağ University under project numbers: 015/3, 016/9. References [1] S. D. Alvarado, A. Dujella, F. Luca, On a conjecture regarding balancing with powers of Fibonacci numbers, Integers, 1 (01), 117 1158. [] T. Andreescu and D. Andrica, Quadratic Diophantine equations, Springer-New York, (015), 14 16. [3] A. Behera, K. Liptai, G. K. Panda, L. Szalay, Balancing with Fibonacci powers, Fibonacci Q., 49 (011), 8 33. [4] A. P. Chaves, D. Marques, A. Togbé, On the sum of powers of terms of a linear recurrence sequence, Bull. Braz. Math. Soc. New Series, 43 (01), 397 406. [5] T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons., 011.

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